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Calc, integration, and me forgetting it

Food?Food? Registered User regular
edited September 2007 in Help / Advice Forum
I'm in Calc 2, and it's been over a year and a half since I had any math class. Thus, I've forgotten even the basics.

What we're working on now is integration by parts. Example problem 1:

integral of (216(x^2)cos(6x))

So what I did so far was set
u=216(x^2)        dv=cos(6x)
du=432x            v=(sin(6x)/6)
I'm pretty sure it's correct so far, but I honestly can't remember how to do it. Anyways, I now have

216(x^2)(sin(6x)/6)- int[(sin(6x)/6)(432x)dx]

Which, when solved, is equal to

36(x^2)(sin(6x))+6(x^2)(cos(6x))

which is wrong. Where did I mess up?

gr_smile2.gif
Food? on

Posts

  • Chief1138Chief1138 Registered User regular
    edited September 2007
    it's been a while for me too, but offhand I'd say switch what you have for u and dv, because it's much easier to find the derivative of cos6x than the antiderivative...also when you find the antiderivative of 216x^2, it reduces to 72x^3 which is a pretty easy number to work with.

    EDIT: given that, let me know if you can work out the rest, if not I'll try to rack my brains for the rest. Looking at it though I think you'll probably have to do IBP once more to get the answer.

    EDIT2: oops maybe I was wrong, this should not be that difficult to solve. Hold on a second.

    EDIT3: OK I'm pretty much stumped here. Too many cobwebs in my brain...Reasonably certain that it's the second part of your solution that is incorrect though. Sorry I couldn't help more.

    Chief1138 on
  • cfgausscfgauss Registered User regular
    edited September 2007
    Are you sure you copied this problem right? It seems weird to have that random 216 in front of everything. Since that's a number, you can just take it out of the integral, since integral[ a f(x) dx ] = a integral[ f(x) dx] (since integrals are just sums).

    In problems like this you want to use u=x^p and dv=trig(x). That way you eliminate powers of x to get something you know how to integrate.

    So here you have u=x^2, dv=cos(6x) dx
    So, du =2x dx, v = 1/6 sin(6x)

    So u*v - Integral [ v du ] = x^2 sin(6x)/6 - Integral[ 2x 1/6 sin(6x) ]

    Then, do integration by parts again to get the integral of cosine, which is something that's easy to do.

    cfgauss on
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  • Food?Food? Registered User regular
    edited September 2007
    cfgauss wrote: »
    Are you sure you copied this problem right? It seems weird to have that random 216 in front of everything. Since that's a number, you can just take it out of the integral, since integral[ a f(x) dx ] = a integral[ f(x) dx] (since integrals are just sums).

    In problems like this you want to use u=x^p and dv=trig(x). That way you eliminate powers of x to get something you know how to integrate.

    So here you have u=x^2, dv=cos(6x) dx
    So, du =2x dx, v = 1/6 sin(6x)

    So u*v - Integral [ v du ] = x^2 sin(6x)/6 - Integral[ 2x 1/6 sin(6x) ]

    Then, do integration by parts again to get the integral of cosine, which is something that's easy to do.

    That's what I did, and WebWork still says it's incorrect. Here's what's been entered as my answer:

    216((x^2)[ sin(6x)/6]+[ x/18]cos(6x))+C

    The parentheses are somewhat confusing because WebWork automatically takes what you've entered and converts it into something it understands before it checks it. Why is that answer wrong?

    Food? on
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  • Food?Food? Registered User regular
    edited September 2007
    Okay, so I had someone from my class show the answer to me, and apparently it's

    36(x^2)sin(6x) +(12x)cos(6x) -(2)sin(6x) +C

    but I can't figure out how. I'd really like to learn where I messed up, so I don't do it again.

    Food? on
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  • inlemurinlemur Registered User regular
    edited September 2007
    I'm leaving out the factor of 216 for simplicity's sake.
    int(x^2*cos(6*x),x) = x^2*sin(6*x)/6-1/6*int(2*x*sin(6*x),x)
    =x^2*sin(6*x)/6-1/3*(-x*cos(6*x)/6+int(cos(6*x)/6,x))
    =x^2*sin(6*x)/6-1/3*(-x*cos(6*x)/6+sin(6*x)/36)
    =x^2*sin(6*x)/6+1/18*x*cos(6*x)-1/108*sin(6*x)
    Multiply through by the 216 and you get the answer.

    inlemur on
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  • Marty81Marty81 Registered User regular
    edited September 2007
    Just throwing this out there: Personally, I can never remember the formula for integration by parts. I always re-derive it when I need it. It comes straight from the product rule,

    (uv)' = uv' + vu'.

    Just integrate to get

    uv = int(uv') + int(vu')

    and rearrange to get

    int(uv') = uv - int(vu').

    Anyway,
    Food? wrote: »
    Anyways, I now have

    216(x^2)(sin(6x)/6) - int[(sin(6x)/6)(432x)dx]

    Which, when solved, is equal to

    36(x^2)(sin(6x)) + 6(x^2)(cos(6x))

    which is wrong. Where did I mess up?

    The bolded parts are not equal. The bolded integral is another integral you need to do...by parts.

    Marty81 on
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